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Let r be the monthly rent per unit in an apartment building with 100 units. All the units are rented at r= $900. One unit becomes vacant for each $10 increase in rent. Maintenance cost are $100 for occupied units and nothing for unoccupied units..... so A) show that the number of units rented is n= 190 - (r/10) for 900 is less than or equal to r and 1900 is greater than or equal to r B) net cash intake= revenue - maintenance costs. determine the rent r that maximizes net cash intake Note: the answer to part A is 0 is less than or equal to r and 100 is greater than or equal to r .. i just need help with part B

  • Calculus - ,

    n=floor(190-(r/10)) for 900≤r≤1900
    Since the number of units has to be integral, and the question says vacancy increases for an increase of $10.
    900≤r≤1900 can also be expressed as an interval [900,1900].

    Revenue for each unit = r-100
    number of units = n = floor(190-(r/10))
    Profit, P(r)
    =Revenue - maintenance
    = (190-(r/10))(r-100)
    = (r^2-2000*r+190000)/10
    = (2000r-r²-190000)/10
    For a maximum profit, set marginal profit to zero
    P'(r) = 200-2r/10 = 0
    if r=1000 is at the maximum, P"(r)<0
    P"(r) = -2/10, therefore r=1000 is a maximum.
    Also check that n is an integral number:
    Profit = P(1000)=$81,000

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